tags: `leetcode` `easy` `Math` `Binary Search`

69. Sqrt(x)

Description

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.

You must not use any built-in exponent function or operator.

For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

Examples

Example 1:

Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.

Example 2:

Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842…, and since we round it down to the nearest integer, 2 is returned.

Constraints:

$0 \leq x \leq 2^{31} - 1$

Code

int mySqrt(int x) {
    int pre = x / 2, now = x / 4;
    if (x == 1)
        return 1;
    // 4 is the largest number that (num^2)/4 <= num
    if (x <= 4)
        return x / 2;
    while (1) {
        // if search window cross the answer, the upper range is the answer
        if (x / pre >= pre)
            return pre;
        // using binary search concept to find the range that sqrt exist
        if (x / now < now || (x / now == now && x % now)) {
            pre = now;
            now /= 2;
        }
        // if we find the range search form both ended
        else {
            pre--;
            now++;
        }
    }
}

Complexity

Space	Time
$O(1)$	$O(log(N))$

Result

Runtime : 36 ms, 17%
Memory usage : 5.7 MB, 31.62%

tags: leetcode easy Math Binary Search